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3 years ago

Can any kind soul help me please

Mathematics

1 answer:

NeX [460]3 years ago

3 0

Answer:

Cot is an inverse of tan:

${ \tt{4 \cot( \theta) = \tan( \theta) }} \\ { \tt{ \frac{4}{ \tan( \theta) } = \tan( \theta) }} \\ \\ { \tt{4 = { \tan }^{2} \theta }} \\ { \tt{ \tan \theta = ±2}} \\ { \bf{\theta =±63.4 \degree }}$

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I need help with sum of my algebra hw assignments

Natalka [10]

Answer:

Step-by-step explanation:

Those lines both make up the tip of the triangle figure...

7 0

3 years ago

Triangles ABC and DFG are similar. Which proportion can be used to find the value of DG¯¯¯¯¯¯? 74=20DG 75=25DG 45=20DG 420=7DG

Tom [10]

The correct answer would be B

5 0

3 years ago

Read 2 more answers

A half-century ago, the mean height of women in a particular country in their 20s was 64.7 inches. Assume that the heights of

Ainat [17]

Answer:

99.5% of all samples of 21 of today's women in their 20's have mean heights of at least 65.86 inches.

Step-by-step explanation:

We are given that a half-century ago, the mean height of women in a particular country in their 20's was 64.7 inches. Assume that the heights of today's women in their 20's are approximately normally distributed with a standard deviation of 2.07 inches.

Also, a samples of 21 of today's women in their 20's have been taken.

Let $\bar X$ = sample mean heights

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean height of women = 64.7 inches

$\sigma$ = standard deviation = 2.07 inches

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the sample of 21 of today's women in their 20's have mean heights of at least 65.86 inches is given by = P( $\bar X$ $\geq$ 65.86 inches)

P( $\bar X$ $\geq$ 65.86 inches) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ $\geq$ $\frac{65.86-64.7}{\frac{2.07}{\sqrt{21} } }$ ) = P(Z $\geq$ -2.57) = P(Z $\leq$ 2.57)

= 0.99492 or 99.5%

The above probability is calculated by looking at the value of x = 2.57 in the z table which has an area of 0.99492.

Therefore, 99.5% of all samples of 21 of today's women in their 20's have mean heights of at least 65.86 inches.

3 0

3 years ago

The graph below shows the distance, y, in miles, of a bird from its nest for a certain amount of time, x, in minutes:

Montano1993 [528]

Answer:

$0.2$ miles per minute represents the speed of the bird and 3 miles represents the original distance of the bird from its nest.

Step-by-step explanation:

As there is no graph mentioned here but the information are quite sufficient to answer the question.

We have points $(0,3), (5,4),(10,5)...$

From these points we can find the slope of the line .

From point slope formula $y-y_1=m(x-x_1)$

And assigning $(x,y) (0,3)$ and $(x_1,y_1) (5,4)$

$m=\frac{y_1-y}{x_1-x} =\frac{4-3}{5-0} =\frac{1}{5}=0.2$

This slope is also the speed of the bird which is $0.2\ miles\ per\ minute.$

As by plugging the values of any coordinate point we can confirm this.

Lets put $(10,5)$ , y-axis is the distance so in $10$ minutes the the distance covered by the bird must be equal to to y-axis value which is $5$ miles.